a) \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)

Từ \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) = k ( k \(\in\) Q, k \(\ne\) 0 )

=> \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

VP = \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2.b.k+3.d.k}{2b+3d}\) = \(\dfrac{k.\left(2b+3d\right)}{2b+3d}\) = k (1)

VT = \(\dfrac{2a-3c}{2b-3d}\) = \(\dfrac{2.b.k-3.d.k}{2b-3d}\) = \(\dfrac{k.\left(2b-3d\right)}{2b-3d}\) = k (2)

Từ (1) và (2) ta có: \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)

hay: (2a+3c).(3b-3d) = (2a-3c).(2b+3d)

b: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=k^2\)

\(\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}=\dfrac{\left(bk-dk\right)^2}{\left(b-d\right)^2}=k^2\)

Do đó: \(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}\)

c: \(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3}{d^3}\)

\(\dfrac{a^3-b^3}{c^3-d^3}=\dfrac{b^3k^3-b^3}{d^3k^3-d^3}=\dfrac{b^3}{d^3}\)

Do đó: \(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{a^3-b^3}{c^3-d^3}\)

d: \(\dfrac{a^{2018}-b^{2018}}{a^{2018}+b^{2018}}=\dfrac{b^{2018}k^{2018}-b^{2018}}{b^{2018}k^{2018}+b^{2018}}=\dfrac{k^{2018}-1}{k^{2018}+1}\)

\(\dfrac{c^{2018}-d^{2018}}{c^{2018}+d^{2018}}=\dfrac{k^{2018}-1}{k^{2018}+1}\)

Do đó: \(\dfrac{a^{2018}-b^{2018}}{a^{2018}+b^{2018}}=\dfrac{c^{2018}-d^{2018}}{c^{2018}+d^{2018}}\)

Cứu mình với 9:00 sáng nay mình nộp bài rùi

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\Rightarrow\left(a+b+c\right)\left(ab+ac+bc\right)-abc=0\Rightarrow\left(a+b\right)\left(ab+ac+bc\right)+abc+ac^2+bc^2-abc=0\Rightarrow\left(a+b\right)\left(ab+ac+bc\right)+c^2\left(a+b\right)=0\Rightarrow\left(a+b\right)\left(ab+ac+bc+c^2\right)=0\Rightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\Rightarrow\left[{}\begin{matrix}a+b=0\\a+c=0\\b+c=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=-b\\c=-a\\b=-c\end{matrix}\right.\)TH1: nếu a=-b

P=(a²⁰¹⁷+b²⁰¹⁷)(b²⁰¹⁸-c²⁰¹⁸)=(-b²⁰¹⁷+b²⁰¹⁷)(b²⁰¹⁸-c²⁰¹⁸)=0

TH2: nếu b=-c

P=(a²⁰¹⁷+b²⁰¹⁷)(b²⁰¹⁸-c²⁰¹⁸)=(a²⁰¹⁷+b²⁰¹⁷)((-c)²⁰¹⁸-c²⁰¹⁸)=0

Còn một TH nữa thì bạn ghi thiếu đề rồi

Giảng cho e vs ak e cần gấp

a: H=5|3x-6|+100>=100

Dấu = xảy ra khi x=2

b: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)

\(\left(\dfrac{a+2018c}{b+2018d}\right)^2=\left(\dfrac{bk+2018dk}{b+2018d}\right)^2=k^2\)

=>ĐPCM

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có

\(VT:\frac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\frac{b^{2018}\cdot k^{2018}+d^{2018}\cdot k^{2018}}{b^{2018}+d^{2018}}=\frac{k^{2018}\left(b^{2018}+d^{2018}\right)}{b^{2018}+d^{2018}}=k^{2018}\)

\(VP:\frac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}=\frac{\left(bk+dk\right)^{2018}}{\left(b+d\right)^{2018}}=\frac{k^{2018}\cdot\left(b+d\right)^{2018}}{\left(b+d\right)^{2018}}=k^{2018}\)

\(\Rightarrow VT=VP\)

Hay \(\frac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\frac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}\left(đpcm\right)\)